# Linear First Order ODE/y' = x + y

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## Theorem

- $(1): \quad \dfrac {\d y} {\d x} = x + y$

has the general solution:

- $y = C e^x - x - 1$

### Initial Value: $\map y 0 = 1$

- $(1): \quad \dfrac {\d y} {\d x} = x + y$

with initial condition:

- $\map y 0 = 1$

has the particular solution:

- $y = 2 e^x - x - 1$

## Proof

Rearranging $(1)$:

- $(2): \quad \dfrac {\d y} {\d x} - y = x$

$(2)$ is a linear first order ODE in the form:

- $\dfrac {\d y} {\d x} + \map P x y = \map Q x$

where:

- $\map P x = -1$
- $\map Q x = x$

Thus:

\(\displaystyle \int \map P x \rd x\) | \(=\) | \(\displaystyle -\int \rd x\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle - x\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle e^{\int P \rd x}\) | \(=\) | \(\displaystyle e^{-x}\) | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac 1 {e^x}\) |

Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:

\(\displaystyle \map {\dfrac {\d} {\d x} } {\dfrac y {e^x} }\) | \(=\) | \(\displaystyle \frac x {e^x}\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \dfrac y {e^x}\) | \(=\) | \(\displaystyle \int \frac x {e^x} \rd x\) | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle -\frac 1 {e^x} \paren {x + 1} + C\) | Primitive of $x e^{a x}$ | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle y\) | \(=\) | \(\displaystyle C e^x - x - 1\) |

$\blacksquare$

## Sources

- 1972: George F. Simmons:
*Differential Equations*... (previous) ... (next): Miscellaneous Problems for Chapter $2$: Appendix $\text{A}$. Numerical Methods